Reading and grades (10.2) Summary statistics for the two groups from Minitab are provided below.

(a) Explain why it is acceptable to use two-sample t procedures in this setting.

(b) Construct and interpret a $95\%$ confidence interval for the difference in the mean English grade for light and heavy readers.

(c) Does the interval in part (b) provide convincing evidence that reading more causes an increase in students’ English grades? Justify your answer.

The figure is,

The numerical summary is,

$\begin{array}{ccccc}\text{Type of reader}& \mathrm{N}& \text{Mean}& \text{StDev}& \text{SE Mear}\\ \text{Heavy}& 47& 3.640& 0.324& 0.047\\ \text{Light}& 32& 3.356& 0.380& 0.067\end{array}$

The formula to compute the confidence interval is:

$\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2,df}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Here following conditions are fulfilled:

a. Samples have been chosen randomly.

b. Normality assumption is also fulfilled because the size of each of the samples is more than $30$.

c. Samples have been chosen independently.

Thus, all the required conditions for the two-sample t-test are fulfilled here.

$\begin{array}{ccccc}\text{Type of reader}& \mathrm{N}& \text{Mean}& \text{StDev}& \text{SE Mear}\\ \text{Heavy}& 47& 3.640& 0.324& 0.047\\ \text{Light}& 32& 3.356& 0.380& 0.067\end{array}$

Using the Ti-83 calculator, the confidence interval is computed as:

The required confidence interval is $(0.11968,0.44832)$.

Interpretation:

There is a $95\%$ probability that difference of population means is between $0.11968$ and $0.44832$.

$\begin{array}{ccccc}\text{Type of reader}& \mathrm{N}& \text{Mean}& \text{StDev}& \text{SE Mear}\\ \text{Heavy}& 47& 3.640& 0.324& 0.047\\ \text{Light}& 32& 3.356& 0.380& 0.067\end{array}$

The provided study is an observational study. So, many confounders may be present which do not provide sufficient evidence to conclude that evidence that reading is leading increase in English's grades.